Absolute minimum of $f(x,y,z)=xy+yz+zx$ on $x^2+y^2+z^2=12$

Question

I'm trying to find the absolute minimum of $f(x,y,z)=xy+yz+zx$ on $g(x,y,z)=0$ where $g(x,y,z)=x^2+y^2+z^2-12$.

I'm using the method of Lagrange multipliers: solving the system $$\nabla f=t\nabla g\\g=0$$

From $y=2t-z$ and $z=2ty-x$ I get $z(2t+1)=x(2t+1)(2t-1)$. Either $t=-1/2$ or $z=x(2t-1)$. Assume the latter. Then again from $y=2tx-z$ I get $y=2tx-x(2t-1)=x$. From $x+y=2tz$, $2x=2tx(2t-1)$. Either $t=0$ or $4t^2-2t-2=0$. The former does not occur because $(0,0,0)$ does not lie on the sphere. Thus the latter occurs. Then either $t=1$ or $t=-1/2$. We've assumed $t\ne -1/2$ so $t=1$. Thus in this case $x=y=0$.

So either $t=-1/2$ or $t=1$. In the former case $x+y+x=0$ and I got using substitutions that the minimum is $-6$. But what to do with $t=1$? Then $f(x,y,z)=3x^2$ and $3x^2=12$ (so $x=\pm 2$)...

Answer 1

Alt. hint: $\;2f(x,y,z)=(x+y+z)^2-(x^2+y^2+z^2) \ge -12\,$.

Answer 2

Calling $M = \left( \begin{array}{ccc} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0 \\ \end{array} \right)$

we have

$$ \min_p p^{\dagger}M p\;\;\;\mbox{s. t.}\;\;\; p^{\dagger}p = 12 $$

This can be solved with Lagrange multipliers so

$$ L(p,\lambda) = p^{\dagger}M p+\lambda(p^{\dagger}p - 12) $$

with the stationary conditions

$$ \nabla L = \left\{ \begin{array}{rcl} 2 \lambda x+y+z&=&0 \\ x+2 \lambda y+z&=&0 \\ x+y+2 \lambda z&=&0 \\ x^2+y^2+z^2-12 &=&0 \\ \end{array} \right. $$

and solving for $p,\lambda$

$$ \left[ \begin{array}{ccccc} x & y & z & \lambda & p^{\dagger}Mp\\ -2 & -2 & -2 & -1 & 12 \\ -2 & 1-\sqrt{3} & 1+\sqrt{3} & \frac{1}{2} & -6 \\ -2 & 1+\sqrt{3} & 1-\sqrt{3} & \frac{1}{2} & -6 \\ 2 & 2 & 2 & -1 & 12 \\ 2 & -1-\sqrt{3} & -1+\sqrt{3} & \frac{1}{2} & -6 \\ 2 & -1+\sqrt{3} & -1-\sqrt{3} & \frac{1}{2} & -6 \\ \end{array} \right] $$

and the minimum for $p^{\dagger}Mp$ is $-6$.

NOTE:

Those points should be qualified with the hessian of $(p^{\dagger}M p)\circ (p^{\dagger}p-12)$

Here from $p^{\dagger}p-12=0\Rightarrow x = \pm\sqrt{12-y^2-z^2}$ and substituting into $p^{\dagger}M p$ after taking the positive option, we obtain

$$ (p^{\dagger}M p)\circ (p^{\dagger}p-12) = \phi(y,z) = z \sqrt{12-y^2-z^2}+y \left(\sqrt{12-y^2-z^2}+z\right) $$

and

$$ H_{\phi} = \{\partial_{\{y,z\}}\partial_{\{y,z\}}\phi(y,z)\} $$

with this hessian we can finally qualify the found stationary points.